Electrochemical

and Photochemical
kinetics
Lecture Notes in Chem. 260
Chemical Kinetics
(Physical Chemistry II)
Joel R. Salazar, Ph.D.
Electrode Kinetics
 Electrochemical Reactions depend on

i.) diffusion of reactants to the surface (electrode)

ii.) reaction on or near the surface
iii.) diffusion of products away from the surface
 Reactants are transported to and products away from an electrode by
three mechanisms:
(1) diffusion
(2) migration, involves the movement of ions through a solution as
a result of electrostatic attraction between the ions and the
electrodes.
(3) convection (as a result of stirring, vibration, or temperature
gradients)
Relation between Current and
Rate
 Consider a surface (e.g.Pt) in contact with a
solution containing M+ and M2+ ion ( as a source and
a sink of electrons)
M2+ + e- ===== M+ (reduction)
M+ ===== M2+ + e- (oxidation)
 The metal surface provides “ sea of electrons”

 The electron conc. does not change with time

Rate of the first reaction: Vc = kc n[M2+]

Rate of the second reaction : Va = kan[M+]
Where n[M2+] and n[M+] = no of moles of ions
Electrochemical Process
 Oxidation = anodic process (a)
 Reduction = cathodic process (c)

 Define: Rate per unit area (mol/m2s) =current

density

 Current is the flux of charges, and thus kinetic

rate and electric current can be equated.
 Current Density
 Current density associated with the first reaction is (reduction)
ic( C/m2s) = z (mol e-/molM2+)F(C/mole e-)Vc(mol M2+/ m2s)

Current density associated with oxidation

ia = zFVa
 Anodic and cathodic currents flow continuously between Pt and the
electrolyte solution
 When equilibrium is established between Pt and the solution, the
cathodic current is equal to and opposite in magnitude to the anodic
current and there is no net current.
io = (ic)o = -( ia)o
inet = (-ia)o + (ic)o
inet = 0
io = exchange current
The Current –Potential Relation:
The Butler –Volmer Equation
 Consider :
i.) dissolution of metal
M(metal) ===== Mz+(aq) + ze- (metal)
ii.) metal –deposition reaction
Mz+(metal) + ze- (metal) === M (metal)
 When equilibrium is established;

Mz+ (metal) === Mz+ (aq)  Charge transfer reaction

 The metal dissolution reaction is one in which a metal ion

(Mz+) (as metal) passes over an activation potential barrier
and becomes a metal ion in the aqueous solution.
The Butler –Volmer Equation
 When no applied potential (φ = 0)
ΔGǂf = for anodic reaction
ΔGǂr = for cathodic reaction
 No electrical difference between the metal and the solution.

• When potential is applied (φ > 0)

 The Gibbs energy of the metal ion in the metal is raised by an amount zFφ
 The Gibbs energy of the metal ion in the activated state is also raised by
an amount
(1 – α) zFφ
Net Result
ΔG‡+ = ΔG‡f - (1 –α) zFφ ( anodic reaction)
ΔG‡ - = ΔG‡r - (1 –α) zFφ + zFφ
ΔG‡ - = ΔG‡r + αzFφ (cathodic reaction)
The Butler –Volmer Equation
 Anodic and Cathodic current densities
i+ = zFkf Cred kf = anodic reaction constant
i- = zFkr Cox kr = cathodic reaction constant
From TST
kf = (kBT/h) e –ΔG‡(+)/RT kr = (kBT/h) e –ΔG‡(-)/RT

Overpotential: η = φ - φo
Assumptions :
i.) If the conc. of metal ion in the metal is constant C red = (Cred)o
ii.) If the solution is stirred well: C ox = (Cox)o
The Butler –Volmer Equation
 Butler – Volmer Equation

i = io ( e(1-α)fη – e- αfη)
where : f = zF/RT
For many reactions,α = 0.5 and for other reactions α
approaches to ½.
i = io ( e 1/2fη – e -1/2fη)
= 2io sinh ( fη)
where : sin h (x) = (ex - e-x) /2
i = 2 io sinh ( fη)
The Low Overpotential Limit
 When the overpotential is so small ( fη) ≤ 1 less than
0.01 V

i = io ( e(1-α)fη – e- αfη) can be expanded into

i = io { 1 + (1- α) fη + …….. - (1 - αfη + ……)}

i = io fη
η = RTi / Fio
The High Potential Limit
1.) the overpotential is large and positive (anodic electrolysis)
i = io ( e(1-α)fη – e- αfη) where e- αfηis negligible.
i = io ( e(1-α)fη

Tafel Plot = the plot of logarithm of the current density against overpotential
In i = In io + (1 – α)fη
2.) the overpotential is large but negative ( cathodic electrolysis)
i = io ( e(1-α)fη – e- αfη) where e ( 1 - α)fηis negligible
i = - ioe- αfη
In (-i) = In io – αfη  Tafel plot
Sample Problem
1.) A typical exchange current density, that for
H+ discharge at platinum is 0.79 mA cm-2 at
25oC. What is the current density at an
electrode when its overpotential is 10 mV.
Take α =0.5
Sample Problem
2.)What overpotential is needed to sustain a
current density of 20 mAcm-2 at a Pt
/Fe3+,Fe2+ electrode in which both ions are at
a mean activity a = 0.10 and the exchange
current density of 2.5 mA cm-2.
 Corrosion
 Corrosion reaction = redox reactions in which a metal is
attacked by some substances in its environment and
converted to an unwanted compound.

 All metals except gold and platinum are thermodynamically

capable of undergoing oxidation in air. However in some
metals, oxidation can result in formation of an insulating
protective oxide layer that prevents further corrosion
passivation layer

 e.g. aluminum, magnesium metal alloys (stainless steel

and nichrome)
 Protective coat of oxide : Al2O3 ( the oxide coat is
impermeable to the passage of O2 and H2O)
 Stainless steel ( iron alloys( 0.4 %C, 18 % Cr and 8 % Ni)
 Metallic Corrosion
Components of an electrochemical corrosion cell
1.) Anode – site at which oxidation occurs (region of high oxidation potential)
Anodic area is usually located at:

1.)stressed region such as portions of large deformation, cracks, scratches, grain boundaries
2.) areas of oxygen starvation
3.)areas of compositional variations
4.) breaks in a protective film

2.) Cathode –site at which reduction occurs

Cathodic area is usually located at
1.) areas of high oxygen concentration
2.) noble metal

3.) electrolyte ( medium through which ions move)

4.) electrical contact (path through which electrons move from anode to cathode)
The thermodynamics of corrosion
 Likelihood of corrosion is obtained by comparing the
standard potentials of the metal reduction
 e.g Fe2+(aq) + 2e- == Fe(s) Eo = -0.44 V
with the values for one of the following half reactions.

For a particular metal

E(M,M+) = Eo (M,M+) + (0.059/Z+) log 10-6
= Eo (M,M+) - (0.35 V/Z+)
Corrosion will occur if

E(H2,H+),, E (O2,H+) and E (O2, OH-) > E (M, M+)

Sample Problem
3 .) Which of the following metals has a
thermodynamic tendency to corrode in moist air at
pH =7; Fe, Cu, Pb. Take as a criterion of corrosion a
metal ion concentration of at least 10-6 mol/L.

Answer: Fe has a thermodynamic tendency to

corrode in moist conditions at pH =7, If oxygen is
absent. But if oxygen is present, all the three
elements have a tendency to corrode.
Corrosion of Iron
 Corrosion of Iron
the rusting of iron requires O2 (iron does not rust in H2O
unless O2 is present)
the rusting of iron also requires H2O ( iron does not rust
in oil, even if it contains O2)
Factors affecting /accelerating rusting
1.) pH of the solution ( high pH but not above pH = 9)
2.) presence of salts
3.) contact with metals more difficult to oxidize than iron
4.) stress on the iron (presence of impurities /possible lattice
defects)
Corrosion of Iron
 Corrosion of iron is believed to be electrochemical in
nature.

 Fe(s) ==== Fe2+ + 2e- Eo(0x) = 0.44 V

O2+4H++ 4e- ===2H2O (l) Eo(red) = 1.23 V

Note : i.) H+ takes part in the reduction of O2

ii.) as pH increases ( H+ is lowered) , the reduction
of O2 becomes less favorable.
Corrosion of iron
 iii.) It is observed that iron in contact with a solution
whose pH is above 9 does not corrode.

The Fe2+ formed at the anode is further oxidized to Fe3+

 4Fe2+ + O (g) + 4H O (l) + 2 xH O == 2Fe O
2 2 2 (l) 2 3
xH2O + 8H+
 The cathode is generally the area having the largest
supply of O2, the rust often deposits there.
Corrosion and…
 Inhibition of Corrosion
1.) coat protective coating
2.) other metal e.g. tin is used to protect its surface against corrosion ( as
long as the protective layer remains intact, once it is exposed to air and
water, the tin actually promotes corrosion of iron
Fe(s)(anode) === Fe2+ + 2e- Eo(ox) = 0.44 v
Sn (s)(cathode) ==== Sn2+ + 2e- Eo (ox) = 0.14 V
3.)Galvanizing (cathodic protection)= protection of a metal by making it
the cathode in an electrochemical cell
e.g. coating iron with a thin layer of zinc (Galvanized iron) =
 The zinc protects the iron against corrosion even after the surface coat is
broken. The iron serves as the cathode in the electrochemical corrosion.
Fe/Fe2+ Eo = 0.44 V Zn/Zn2+ = 0.76 V
 Sacrificial anode
4.) Cathodic Protection by impressed current (electrons are supplied
from an external cell so that the object itself is not oxidized)
 The current provides power source to produce electrons (this method is
used when the replacement of anode is difficult)
…Corrosion Prevention
 Rate of Uniform Corrosion
 The amount of metal uniformly corroded from an anode or
electroplated on a cathode in an aqueous solution can be
determined by using Faraday’s Equation.

 W = ItM/nF w = weight of metal corrode

I = current flow
n = number of electrons /atom
F = 96,500C/mole = As/mol
Sometimes the uniform aqueous corrosion of a metal is
expressed in terms of a current density i ( A/cm2)
W = iAtM/nF
Sample Problem
4.)A copper electroplating process
utilizes15A of current by chemically
dissolving (corroding) a copper anode and
electroplating copper cathode. If it is
assumed that there are no side reactions,
how long will it take to corrode 8.50 g of
copper from the anode?

Answer: 28.7 minutes

Sample Problem
5.) A sample of zinc corrodes uniformly with a
current density of 4.27 x10-7 A/cm2 in an
aqueous solution. What is the corrosion rate
of the zinc in mdd(mg/dm2in1day)
Photochemistry and
Photochemical Reactions
 The interaction of light with matter
 Why important? Life itself depends on
photochemical processes

 e.g. Photosynthesis = energy from the sun

harnessed by living organisms
 Earth’s atmosphere = supports life and
shields us from damaging UV radiation
Photochemistry

What chemicals exist in the earths atmosphere and what chemical

processes can occur?

“Normal” Gases Bio. Processes Atmos. Chem. Anthropogenic

N2 CH3Cl NO
O2 CH4 N2O CFC’s
N2O4 +…
H2O H2S
CO2 NaCl O3
Ar Soot H2SO4

ClO
OH
O
Cl
Photochemistry
The Chemistry of the Earth’s Atmosphere.
Photochemistry

The Chapman Model:

O2 + hv → O + O
185 nm <  < 220 nm Initiation

O3 + hv → O2 + O 210 nm <  < 300 nm

Propagation
O + O2 + M → O3 + M Hr = -106.6 kJ mol -1

O + O3 → 2O2 Hr = -391.9 kJ mol-1

Termination
O + O + M → O2 + M
Photochemistry

A cause of concern is the effect of anthropogenic pollutants

on the ozone concentration.

CFC’s may be photolysed:

CF2Cl2 + hv → CF2Cl· + Cl·

Cl· + O3 == ClO· + O2

ClO· + O == O2 + Cl·

A reduction in stratospheric ozone will cause an increase in

the amount of UV radiation reaching the earth’s surface.
Photochemistry

Global warming (for):

•Rise in CO2 and other

greenhouse gases

•Historical temp. record shows

an increase of 0.4 – 0.8 oC over
the last 100 years.

•Unusual warmth over last 1000

years.
•Climate models can reproduce
•CO2 conc. is the most
warming trend but only when
important indicator of climate
greenhouse gases are included.
change.
Photochemical Principles
1.) Photosynthesis (light is used to bring about
a chemical change
2.) Chemiluminescence (energy released by
chemical reactions producing light
3.) Flourescence/Phosphorescence (physical
processes that do not involve a change in
chemical identity)
 Essential Features of
Photochemistry
The chemistry of excited states ( electron of an
atoms/molecules are distributed in a higher energy
configuration.
Photobiology Spectroscopy Photophysics

Photochemistry

Synthesis Reaction Kinetics Quantum theory

Fate of Excited State (Primary
Photochemical Processes

 Dissociation
 Direct reaction
 Isomerization
 Intermolecular energy transfer
 Intramolecular energy transfer
 Luminescence
 Physical Quenching
 Ionization
Atomic and Molecular Energy
Levels
 Rotational/ Vibrational = arises from periodic
oscillation of atoms

 Electronic = depends on electron position

(distance from the nucleus)

 Only specific frequencies of radiation can be

absorbed or emitted ( internal energy levels in
atoms and molecules are quantized)
Photochemistry – some background

At very short internuclear distances

the atoms repel one another

At large internuclear
distancesthe atoms
cannot attract one
another. They are
separated.
Absorption of Radiation
 Absorption of light usually occurs in a single step.
 Consider two states L and U.
Bohr Condition EL – Eu =hv.
Note: very high intensity light can produce multiple photon
absorption = Lasers
 Photochemistry involves radiation between 2000 nm (near IR) -
<100 nm (soft x-ray)
 The most important regions
700 – 400 nm (visible)
400 – 200 nm (UV)
200 – 100 nm (VUV)
 Same order as bond energies ( Photochemical Dissociation)
 Also covers the range of activation energies ( Photochemical
reaction)
Emission of Radiation
 Stimulated Emission (induced)
 Exact analogue of absorption
 An excited species interacts with the oscillating
electric field and gives up its energy to the
incident radiation.
 Essential part of laser action
Emission of Radiation
 Spontaneous Emission
 The excited species gives up its energy of its own
accord
(in the absence of a radiation field)
 e.g fluorescence, phosphorescence and
chemiluminescence.
Flourescence and Phosphorescence
The Timescales of Photophysical Processes
•It takes about 10-16 seconds for absorption of UV and visible radiation to take place.

•So, the upper limit for a rate constant for a first-order photochemical reaction is about
1016 s-1.

•Fluorescence is slower than absorption but still very fast and can be used to initiate very
fast reactions:

*
+ hv 

200 fs

0.25 ms
“Biological cascade”

•Phosphorescence is typically much slower than fluorescence.

The Fluorescent Light

Sealed glass tube, electrodes at each end, filled

with argon and a little mercury at very low
pressure. Glass coated with a phosphor
powder.

When we turn on the AC power electrons flow

from one electrode to the other electrode.

Hg + e-  {Hg-}*
{Hg-}* + M  Hg + e- +M* + hv (UV light)

hv (UV) + Phosphor powder  Phosphor powder*

Phosphor powder*  Phosphor powder + hv (visible)
Laws of Photochemistry

 1st law (Grotthus- Draper)

 Only light that is absorbed by a molecule can be
effective in producing photochemical change in that
molecule.
 K2Cr2O7 absorbs light at 440 nm ( violet region of the
spectrum)
 The amount of light absorbed: Ia = Io – It
Io = incident light
It = transmitted light
Laws of Photochemistry
From Beer’s Law
It = Ioe-εcl
log (It/Io) = - εcl
A = εcl

 For a photochemical reaction to occur, there must be

an interaction between radiation and the reactant.
 E = NA hv = NAhc/λ
Laws of Photochemistry
2nd Law (Stark –Einstein)
One particle is excited for each quantum of radiation
absorbed (assuming light intensity is not too high)

(with high intensity laser light, a molecule can absorb

several photons in a single photochemical process =
laser action, exception to the Stark- Einstein Law)

These lead to the concept of Quantum yield.

Photochemistry – The Primary Quantum Yield

In keeping with the timescales of photochemical events there are two processes
classifications:

Primary processes: Products formed directly from excited state

(photoisomerization of retinal)

Secondary processes: Products formed from intermediates which have

been formed directly from the excited state of the reactant.

For a primary process we may define the primary quantum yield as:

number of events rate of process v

  
number of photons absorbed intensity of light absorbed I abs
The Mechanism of Decay of Excited Singlet States.
Consider the formation and decay of an excited singlet state
Rate Law:
Absorption: S + hvi  S* vabs = Iabs
Fluorescence: S*  S + hvf vf = kf[S*]
Internal Conversion: S*  S vic = kic[S*]
Intersystem Crossing: S*  T* visc = kisc[S*]

We find that the quantum yield of fluorescence is:

kf
f 
kf  kic  kisc
Lifetime of flourescence is; τS = 1/( kF + kIC + kISC)

ΦF = τSkF
Quantum yield of Phosphorescence
 Since triplet state molecules may have long lifetimes
compared with singlet state molecules, therefore a
higher probability of undergoing chemical reaction in
phosphorence
 Lifetime of phosphorescence;
τT = 1/ (kP + kISC)
ΦP= rate of phosphorescence emission
rate of absorption of radiation
= kP[T1] / Ia
Ia = kIC[S] + kF[S] + kISC[S]
[S] = Ia / kIC + kF + kISC

The steady state equation for [T]

kISC [S] = kISC’ [T] + kP[T]
[T] = kISC[S] / kISC’ + kP
[T] = kISC Ia / (kIC + kF + kISC) (kISC’ + kP)

ΦP = kPkISC / (kIC + kF + kISC) (kISC’ + kP)

= kP kISC τS τT ( without quencher)
Electronic Energy Transfer in
the presence of quencher (Q)
 Generally observed when one excited molecule undergoes chemical
reaction with another molecule. This reaction is accompanied by
deactivation of the excited molecule, and the excited state is said
to be quenched.
 In the presence of quencher,

[T] = kISC[S] / kISC’ + kP + kQ [Q]

without quencher; [T]o = kISC[S] / kISC’ + kP
By getting their ratios; we get:
kISC’ + kP + kQ [Q]
------------------------- = 1 + kQτT[Q]
kISC’ + kP

Iop / IP = 1 + kQτT[Q] = Stern –Volmer equation

Quantum Yield

 Efficiency of quantum usage

 The Φ (quantum yield) is the number of molecules of reactant
consumed or number of molecules of product produced per
photon/ einstein of light absorbed.
Φ can range from nearly zero to about 106.

A value of φ <1 suggests primary processes, φ >1 suggests

secondary processes.
Φ (overall quantum yield = reflects primary and secondary
processes)
Quantum Yield
 Φ>1 (if secondary reactions occur)

 e.g. HI + hv === H· + I· ( primary process)

H· + HI === H2 + I· (secondary process)
I· + I· === I2 (secondary process)
----------------------------------------
2HI ===== H2 + I2
Φ = 2 molecules of HI/ 1 photon
absorbed
Sample problem
1.) In the photobromination of cinnamic acid,
radiation at 435.8 nm with an intensity of 1.43
x10-3J/s was 80.1 % absorbed in a li of
solution during an exposure of 1105 sec. The
conc. of Br2 decreased by 7.5 x 10-5 mol/L
during this period. What is the quantum yield?

Ans. 16.6
Sample Problem
2.) When acetone was photolyzed at 56oC with 313 nm radiation for
23,000 sec. 5.23 x 1019 molecules were decomposed. If 8.52 x 10-
3
J of radiation were absorbed per second. Calculate the quantum
yield.

Answer : 0.17

3.) For the reaction

A + 2B === C, express d[A]/dt, d[B]/dt and d[C]/dt in terms
of Ia andsφ.
-d[A]/dt = φIa
-1/2 d[B]/dt = φIa = 2 φIa
d[C]/dt = φIa
Actinometry (Physical
Devises)
 Primary Standards
 Thermal detectors = converts incident photons
to heat and measure the heat released (requires
monochromatic light)
 Quantum counters = uses a dye, normally
rhodamine B, to absorb the photons and emit
flourescence, eventually detected with a
photomultiplier
Actinometry (chemical
systems)
 Chemical Systems
 Chemical systems for which the quantum yield of a
given process is accurately known, and thus can be
employed as a standard for the determination of the
photon flux.
 Wavelength, temperature and reactant concentration
are dependent and affected by impurities
e.g. Potassium Ferrioxalate K3Fe (C2O4)3
converts to Fe2+ in acid solution
Uranyl oxalate
Azobenzene and its derivatives
Photochemical Kinetics
 The rate of the primary activating process is controlled by
the intensity (Ia) of the activating radiation used and
proportional to the intensity of radiation.

 Consider a hypothetical reaction

A2 ==== 2A
(proceeds by photochemical activation)
Mechanism
A2 + hv === A2* k1 (activation)
A2* === 2A k2 (dissociation)
A2* + A2 === 2A2 k3 ( deactivation)